弹出后显示单击按钮不起作用?

时间:2015-01-12 13:46:04

标签: android popupwindow

当显示弹出窗口时,单击按钮不能显示哪个弹出窗口?

private void initiatePopupWindow(ScrollView sv) {
    try {


       if(popOpened){

            pwindo.dismiss();
        popOpened = false;
       }
        else{

        pwindo = new PopupWindow(sv, LayoutParams.WRAP_CONTENT, 800, true);
        pwindo.showAsDropDown(filterButton,30,30);
         popOpened = true;
      }


    } catch (Exception e) {
        e.printStackTrace();
    }
}

filterButton.setOnClickListener(new View.OnClickListener(){

        @Override
        public void onClick(View arg0) {
            System.out.println("this is click of filter button:::");



            if (filterProduct.isEmpty()) {

                GetFilterWS task = new GetFilterWS();
                task.execute();
            } else {
                initiatePopupWindow(sv);

            }

        }
    });

显示弹出窗口时不打印System.out.println消息。

1 个答案:

答案 0 :(得分:0)

您是否尝试过PopupWndow.setOutsideTouchable(true);